# pink_and_blue

SSkyFire
java
2 months ago
5.3 kB
0
Indexable
Never
```Pink and Blue
Xenny was a teacher and he had N students. The N children were sitting in a room. Each child was wearing a white T-shirt, with a unique number from the range 1 to N written on it. T-Shirts of pink and blue color were to be distributed among the students by Xenny. This made the students very happy.

Xenny felt that a random distribution of T-Shirts would be very uninteresting. So, he decided to keep an interesting condition:

Every student would get a T-Shirt that is of a different color than his/her friends. That is, if X and Y are friends and X has a Pink T-Shirt, then Y should compulsorily have a Blue T-Shirt, and vice-versa.

Also, Xenny had a belief that Boys should wear blue T-Shirts and Girls should wear pink T-Shirts. If a boy was given a pink T-Shirt or a girl was given a Blue T-Shirt, he called it an inversion.

So, Xenny wanted to distribute T-Shirts in the above-mentioned interesting manner and also wanted to minimize "inversions". Help him solve the task.

Note: There are no disjoint groups of friends in the room. That is, 2 distinct groups with finite number of students do not exist, but exactly 1 group of students exists in the given situation.

Input
The first line is the number of test cases T.

First line of each test case contains 2 space-separated integers - N and M - number of students and number of friendships present respectively.

Second line consists of N space-separated characters, where ith character denotes the gender of the ith student. B: Boy, G: Girl.

M lines follow. Each line consists of 2 space-separated integers, u and v, showing that u is a friend of v and vice-versa.

Output
If Xenny could distribute the T-Shirts in the desired way, print the minimum number of inversions required.
Else, print -1.

Constraints
1 ≤ N ≤ 105
1 ≤ M ≤ 105
1 ≤ u, v ≤ N

Colors of T-Shirt are represented by uppercase characters 'B' and 'G'

Sample

Input

3

3 2

B G B

1 2

1 3

6 9

B B B G G G

3 5

2 6

4 2

6 3

3 1

3 4

6 1

5 1

1 4

6 5

G G G B G G

6 3

1 3

2 3

4 3

5 3

Output

1

-1

2

Explanation

#1

Student 1 can be given a Blue T-Shirt. Hence, Student 2 and 3 would receive Pink T-Shirts. Since, Student 3 is a Boy and has received a Pink T-Shirt, number of inversions = 1.

package blue_and_pink;

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class Solution {
static int n, friendship;
static char []a;
static int []color;
static int [][]map;
public static class Queue{
final int MAX_SIZE = 1000000;
int []data = new int[MAX_SIZE];
int front;
int rear;
public Queue() {
front = rear = -1;
}
public void init() {
front = rear = -1;
}
public boolean isEmpty() {
if(front == rear) return true;
else return false;
}
public void enQueue(int element) {
rear = rear + 1;
data[rear] = element;
}
public int deQueue() {
front = front + 1;
return data[front];
}
}

static Queue queue;

public static void resetColor() {
for(int i=0; i<n; i++) {
color[i] = 0;
}
}
//1: xanh, 2 hong
public static int bfs(int start, int mau) {
queue.init();
resetColor();
queue.enQueue(start);
color[start] = mau;
boolean check = true;
while (!queue.isEmpty()) {
int p = queue.deQueue();
for(int i=0; i<friendship; i++) {
if(map[i][0] == p) {
if(color[map[i][1]] == 0) {
color[map[i][1]] = 3 - color[p];
queue.enQueue(map[i][1]);
}
else if(color[map[i][1]] == color[p]) {
check = false;
}
}
else if(map[i][1] == p) {
if(color[map[i][0]] == 0) {
color[map[i][0]] = 3 - color[p];
queue.enQueue(map[i][0]);
}
else if(color[map[i][0]] == color[p]) {
check = false;
}
}
}
if(check == false) {
break;
}
}

if(check == true) {
int sumInversion = 0;
for(int i=1; i<=n; i++) {
if(a[i] == 'B' && color[i] != 1 || a[i] == 'G' && color[i] != 2) {
sumInversion ++;
}
}
return sumInversion;
}
else {
return -1;
}
}

public static void main(String[] args) {
try {
System.setIn(new FileInputStream("src/blue_and_pink/input.txt"));
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Scanner sc = new Scanner(System.in);
int TC = sc.nextInt();
queue = new Queue();
for(int t=0; t<TC; t++) {
n = sc.nextInt();
friendship = sc.nextInt();
a = new char[n+1];
color = new int[n+1];
map = new int[friendship][2];

for(int i=1; i<=n; i++) {
a[i] = sc.next().charAt(0);
}

for(int i=0; i<friendship; i++) {
map[i][0] = sc.nextInt();
map[i][1] = sc.nextInt();
}

int minInversion = Integer.MAX_VALUE;
for(int i=1; i<=n; i++) {
for(int j=1; j<=2; j++) {
int res = bfs(i, j);
if(res != -1 && res < minInversion) {
minInversion = res;
}
}
}
if(minInversion == Integer.MAX_VALUE) {
System.out.println(-1);
}
else {
System.out.println(minInversion);
}

}
}
}
```